Raising the bar

Properties of root subspaces

Let $latex A$ be a square matrix and let $latex \lambda \in \sigma (A)$ be its eigenvalue. As we know, the nonzero elements of the null space $latex N(A-\lambda I)=\{x:(A-\lambda I)x=0\}$ are the corresponding eigenvectors. This definition is generalized as follows.

Definition 1. The subspaces $latex N_{\lambda }^{(k)}=N((A-\lambda I)^{k}),$ $latex k=1,2,...$ are called root subspaces of $latex A$ corresponding to $latex \lambda .$

Exercise 1. a) Root subspaces are increasing:

(1) $latex N_{\lambda }^{(k)}\subset N_{\lambda }^{(k+1)}$ for all $latex k\geq 1$

and b) there is such $latex p\leq n$ that all inclusions (1) are strict for $latex k<p$ and

(2) $latex N_{\lambda }^{(p)}=N_{\lambda }^{(p+1)}=...$

Proof. a) If $latex x\in N_{\lambda }^{(k)}$ for some $latex k,$ then $latex (A-\lambda I)^{k+1}x=(A-\lambda I)(A-\lambda I)^{k}x=0,$ which shows that $latex x\in N_{\lambda }^{(k+1)}.$

b) (1) implies $latex \dim N_{\lambda }^{(k)}\leq \dim N_{\lambda }^{(k+1)}.$ Since all root subspaces are contained in $latex C^{n},$ there are $latex k$ such that $latex N_{\lambda }^{(k)}=N_{\lambda }^{(k+1)}.$ Let $latex p$ be the smallest such $latex k.$ Then all inclusions (1) are strict for $latex k<p.$

Suppose $latex N_{\lambda}^{(k+1)}\setminus N_{\lambda }^{(k)}\neq \varnothing $ for some $latex k\ge p.$ Then there exists $latex x\in N_{\lambda }^{(k+1)}$ such that $latex x\notin N_{\lambda}^{(k)}$, that is, $latex (A-\lambda I)^{k+1}x=0,$ $latex (A-\lambda I)^{k}x\neq 0.$ Put $latex y=(A-\lambda I)^{k-p}x.$ Then $latex (A-\lambda I)^{p+1}y=(A-\lambda I)^{k+1}x=0,$ $latex (A-\lambda I)^{p}y=(A-\lambda I)^{k}x\notin 0.$ This means
that $latex y\in N_{\lambda }^{(p+1)}\setminus N_{\lambda }^{(p)}$ which contradicts the definition of $latex p.$

Definition 2. Property (2) can be called stabilization. The number $latex p$ from (2) is called a height of the eigenvalue $latex \lambda$.

Exercise 2. Let $latex \lambda \in \sigma (A)$ and let $latex p$ be the number from Exercise 1. Then

(3) $latex C^{n}=N_{\lambda }^{(p)}\dotplus \text{Img}[(A-\lambda I)^{p}].$

Proof. By the rank-nullity theorem applied to $latex (A-\lambda I)^{p}$ we have $latex n=\dim N_{\lambda }^{(p)}+\dim \text{Img}[(A-\lambda I)^{p}].$ By Exercise 3, to prove (3) it is sufficient to establish that $latex L\equiv N_{\lambda}^{(p)}\cap \text{Img}[(A-\lambda I)^{p}]=\{0\}.$ Let's assume that $latex L$ contains a nonzero vector $latex x.$ Then we have $latex x=(A-\lambda I)^{p}y$ for some $latex y.$ We obtain two facts:

$latex (A-\lambda I)^{p}y\neq 0$ $latex \Longrightarrow y\notin N_{\lambda }^{(p)},$

$latex (A-\lambda I)^{2p}y=(A-\lambda I)^{p}(A-\lambda I)^{p}y=(A-\lambda I)^{p}x=0\Longrightarrow y\in N_{\lambda }^{(2p)}.$

It follows that $latex y$ is a nonzero element of $latex N_{\lambda }^{(2p)}\setminus N_{\lambda }^{(p)}.$ This contradicts (2). Hence, the assumption $latex L\neq \{0\} $ is wrong, and (3) follows.

Exercise 3. Both subspaces at the right of (3) are invariant with respect to $latex A.$

Proof. If $latex x\in N_{\lambda }^{(p)},$ then by commutativity of $latex A$ and $latex A-\lambda I$ we have $latex (A-\lambda I)^{p}Ax=A(A-\lambda I)^{p}x=0,$ so $latex Ax\in N_{\lambda }^{(p)}.$

Suppose $latex x\in \text{Img}[(A-\lambda I)^{p}],$ so that $latex x=(A-\lambda I)^{p}y$ for some $latex y.$ Then $latex Ax=(A-\lambda I)^{p}Ay\in \text{Img}[(A-\lambda I)^{p}].$

Exercise 3 means that, for the purpose of further analyzing $latex A,$ we can consider its restrictions onto $latex N_{\lambda }^{(p)}$ and $latex \text{Img}[(A-\lambda I)^{p}].$

Exercise 4. The restriction of $latex A$ onto $latex N_{\lambda }^{(p)}$ does not have eigenvalues other than $latex \lambda .$

Proof. Suppose $latex Ax=\mu x,$ $latex x\neq 0,$ for some $latex \mu .$ Since $latex x\in N_{\lambda }^{(p)},$ we have $latex (A-\lambda I)^{p}x=0.$ Then $latex (A-\lambda I)x=(\mu -\lambda )x$ and $latex 0=(A-\lambda I)^{p}x=(\mu -\lambda )^{p}x$. This implies $latex \mu =\lambda .$

Exercise 5. The restriction of $latex A$ onto $latex \text{Img}[(A-\lambda I)^{p}]$ does not have $latex \lambda $ as an eigenvalue (so that $latex A-\lambda I$ is invertible).

Proof. Suppose $latex x\in \text{Img}[(A-\lambda I)^{p}]$ and $latex Ax=\lambda x,$ $latex
x\neq 0.$ Then $latex x=(A-\lambda I)^{p}y$ for some $latex y\neq 0$ and $latex 0=(A-\lambda
I)x=(A-\lambda I)^{p+1}y.$ By Exercise 1 $latex y\in N_{\lambda }^{(p+1)}=N_{\lambda }^{(p)}$ and $latex x=(A-\lambda I)^{p}y=0.$ This contradicts the choice of $latex x.$

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